LeetCode 0496. Next Greater Element I Solution in Java, Python, C++, JavaScript, Go & Rust | Explanation + Code

Description

The next greater element of some element x in an array is the first greater element that is to the right of x in the same array.

You are given two distinct 0-indexed integer arrays nums1 and nums2, where nums1 is a subset of nums2.

For each 0 <= i < nums1.length, find the index j such that nums1[i] == nums2[j] and determine the next greater element of nums2[j] in nums2. If there is no next greater element, then the answer for this query is -1.

Return an array ans of length nums1.length such that ans[i] is the next greater element as described above.

 

Example 1:

Input: nums1 = [4,1,2], nums2 = [1,3,4,2]
Output: [-1,3,-1]
Explanation: The next greater element for each value of nums1 is as follows:
- 4 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1.
- 1 is underlined in nums2 = [1,3,4,2]. The next greater element is 3.
- 2 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1.

Example 2:

Input: nums1 = [2,4], nums2 = [1,2,3,4]
Output: [3,-1]
Explanation: The next greater element for each value of nums1 is as follows:
- 2 is underlined in nums2 = [1,2,3,4]. The next greater element is 3.
- 4 is underlined in nums2 = [1,2,3,4]. There is no next greater element, so the answer is -1.

 

Constraints:

• 1 <= nums1.length <= nums2.length <= 1000
• 0 <= nums1[i], nums2[i] <= 104
• All integers in nums1 and nums2 are unique.
• All the integers of nums1 also appear in nums2.

 

Follow up: Could you find an O(nums1.length + nums2.length) solution?

Solutions

Solution 1: Monotonic Stack

We can traverse the array nums2 from right to left, maintaining a stack stk that is monotonically increasing from top to bottom. We use a hash table d to record the next greater element for each element.

When we encounter an element x, if the stack is not empty and the top element of the stack is less than x, we keep popping the top elements until the stack is empty or the top element is greater than or equal to x. At this point, if the stack is not empty, the top element of the stack is the next greater element for x. Otherwise, x has no next greater element.

Finally, we traverse the array nums1 and use the hash table d to get the answer.

The time complexity is O(m + n), and the space complexity is O(n). Here, m and n are the lengths of the arrays nums1 and nums2, respectively.

PythonJavaC++GoTypeScriptRustJavaScript

class Solution: def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]: stk = [] d = {} for x in nums2[::-1]: while stk and stk[-1] < x: stk.pop() if stk: d[x] = stk[-1] stk.append(x) return [d.get(x, -1) for x in nums1](code-box)

class Solution { public int[] nextGreaterElement(int[] nums1, int[] nums2) { Deque<Integer> stk = new ArrayDeque<>(); int m = nums1.length, n = nums2.length; Map<Integer, Integer> d = new HashMap(n); for (int i = n - 1; i >= 0; --i) { int x = nums2[i]; while (!stk.isEmpty() && stk.peek() < x) { stk.pop(); } if (!stk.isEmpty()) { d.put(x, stk.peek()); } stk.push(x); } int[] ans = new int[m]; for (int i = 0; i < m; ++i) { ans[i] = d.getOrDefault(nums1[i], -1); } return ans; } }(code-box)

class Solution { public: vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) { stack<int> stk; unordered_map<int, int> d; ranges::reverse(nums2); for (int x : nums2) { while (!stk.empty() && stk.top() < x) { stk.pop(); } if (!stk.empty()) { d[x] = stk.top(); } stk.push(x); } vector<int> ans; for (int x : nums1) { ans.push_back(d.contains(x) ? d[x] : -1); } return ans; } };(code-box)

func nextGreaterElement(nums1 []int, nums2 []int) (ans []int) { stk := []int{} d := map[int]int{} for i := len(nums2) - 1; i >= 0; i-- { x := nums2[i] for len(stk) > 0 && stk[len(stk)-1] < x { stk = stk[:len(stk)-1] } if len(stk) > 0 { d[x] = stk[len(stk)-1] } stk = append(stk, x) } for _, x := range nums1 { if v, ok := d[x]; ok { ans = append(ans, v) } else { ans = append(ans, -1) } } return }(code-box)

function nextGreaterElement(nums1: number[], nums2: number[]): number[] { const stk: number[] = []; const d: Record<number, number> = {}; for (const x of nums2.reverse()) { while (stk.length && stk.at(-1)! < x) { stk.pop(); } d[x] = stk.length ? stk.at(-1)! : -1; stk.push(x); } return nums1.map(x => d[x]); }(code-box)

use std::collections::HashMap; impl Solution { pub fn next_greater_element(nums1: Vec<i32>, nums2: Vec<i32>) -> Vec<i32> { let mut stk = Vec::new(); let mut d = HashMap::new(); for &x in nums2.iter().rev() { while let Some(&top) = stk.last() { if top <= x { stk.pop(); } else { break; } } if let Some(&top) = stk.last() { d.insert(x, top); } stk.push(x); } nums1 .into_iter() .map(|x| *d.get(&x).unwrap_or(&-1)) .collect() } }(code-box)

/** * @param {number[]} nums1 * @param {number[]} nums2 * @return {number[]} */ var nextGreaterElement = function (nums1, nums2) { const stk = []; const d = {}; for (const x of nums2.reverse()) { while (stk.length && stk.at(-1) < x) { stk.pop(); } d[x] = stk.length ? stk.at(-1) : -1; stk.push(x); } return nums1.map(x => d[x]); };(code-box)